Question

$\int \frac{2\cos 2x+{\sec }^{2}x}{\sin 2x+\tan x-5}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{2\cos 2x+{\sec }^{2}x}{\sin 2x+\tan x-5}dx \\ \mathrm{Putting}\sin 2x+\tan x-5=t \\ \Rightarrow 2\cos 2x+{\sec }^{2}x=\frac{dt}{dx} \\ \Rightarrow \left(2\cos 2x+{\sec }^{2}x\right)dx=dt \\ \therefore I=\int \frac{1}{t}dt \\ =\ln \left|t\right|+\mathrm{C} \\ =\ln \left|\sin 2x+\tan x-5\right|+\mathrm{C}\left[\because \mathit{}t=\sin 2x+\tan x-5\right] \end{gathered}$$